BSEAP Board Class 10 Mathematics Question Paper 2023 with Answer Key

The prime factorization of the numbers are: \[12 = 2^2 \times 3,\quad 15 = 3 \times 5,\quad 21 = 3 \times 7\]

Thus, the LCM = \(2^2 \times 3 \times 5 \times 7 = 420\).
Quick Tip: To find LCM, always take the highest power of all prime factors involved.

The roster form of the set clearly lists all elements explicitly: \[A = \{1, 2, 3, 4, 5\}\] Quick Tip: Roster form explicitly lists all elements of a set without repetition.

Statement (P) is correct because, by definition, a quadratic polynomial always has a degree equal to 2. Statement (Q) is also correct because the maximum number of zeroes (roots) of any polynomial equals its degree, and hence, a quadratic polynomial can have a maximum of 2 zeroes. Therefore, both statements (P) and (Q) are correct.
Quick Tip: Always remember: the degree of a polynomial determines the maximum possible number of its zeroes.

Both given equations are proportional, as: \[ \frac{3}{1} = \frac{6}{2} = \frac{3900}{1300} = 3 \]

Thus, the lines represented by these equations coincide and have infinitely many solutions. Therefore, both Assertion and Reason are correct, and Reason correctly explains the Assertion.
Quick Tip: If two linear equations have proportional coefficients, they represent coincident lines with infinitely many solutions.

Given the quadratic equation: \[ 5x^2 - 6 = 0 \quad \Rightarrow \quad x^2 = \frac{6}{5} \]

Taking square roots on both sides, we get two solutions: \[ x = \sqrt{\frac{6}{5}} \quad and \quad x = -\sqrt{\frac{6}{5}} \]

Hence, the equation has exactly \(2\) roots.
Quick Tip: A quadratic equation always has exactly two roots, which may be real or imaginary depending on the discriminant.

Thales’ theorem states that if a line is drawn parallel to one side of a triangle, intersecting the other two sides at distinct points, then it divides those sides in the same ratio.
Quick Tip: Thales’ theorem is useful for solving problems involving proportional segments in triangles.

At each endpoint of the diameter of a circle, exactly two tangents can be drawn, which are perpendicular to each other. Therefore, the total number of tangents at the endpoints of the diameter is 2.
Quick Tip: Tangents drawn at the endpoints of a diameter are always perpendicular.

The volume of a cube is given by the formula: \[ V = s^3 \]
Substituting \(s = 4\) cm: \[ V = 4^3 = 64 cm^3 \]
Thus, the volume of the cube is 64 cm³.
Quick Tip: The volume of a cube is always the cube of its side length.

From trigonometric identities: \[ \sin \theta = \frac{\sec^2 \theta -1}{\sec^2 \theta} \] \[ \cos \theta = \frac{1}{\sec \theta} \] \[ \tan \theta = \sqrt{\sec^2 \theta -1} \]
Thus, the correct matching is: \[ P \rightarrow (iii), \quad Q \rightarrow (i), \quad R \rightarrow (ii) \]
which corresponds to option (B).
Quick Tip: Memorizing standard trigonometric identities helps in solving such matching problems quickly.

The given problem represents a right-angled triangle where:
- The observer's position forms the base (\(AB = 20\) m).
- The height of the building is to be determined (\(BC\)).
- The angle of elevation is \(60^\circ\).

By using the formula: \[ \tan 60^\circ = \frac{opposite}{adjacent} = \frac{BC}{AB} \] \[ \sqrt{3} = \frac{BC}{20} \] \[ BC = 20\sqrt{3} \approx 34.64 m \]

Thus, the height of the building is approximately 34.64 meters.
Quick Tip: Use trigonometric ratios like \(\tan\), \(\sin\), or \(\cos\) to solve problems involving heights and distances.

By the probability rule: \[ P(\neg E) = 1 - P(E) \] \[ P(\neg E) = 1 - 0.05 = 0.95 \]

Thus, the probability of not \(E\) is 0.95.
Quick Tip: The sum of the probability of an event and its complement is always 1.

The mean is calculated as: \[ Mean = \frac{\sum data values}{total number of values} \]

Summing the values: \[ 2 + 3 + 7 + 6 + 6 + 3 + 8 = 35 \]

Total number of values = 7.
\[ Mean = \frac{35}{7} = 5 \]

Thus, the mean of the given data is 5.
Quick Tip: The mean (average) is obtained by dividing the sum of all values by the total count.

The intersection of two sets, \( A \cap B \), includes elements that are common in both sets: \[ A \cap B = \{5, 6\} \]

A Venn diagram would represent \( A \) and \( B \) as overlapping circles, with \( \{5,6\} \) in the intersection region.
Quick Tip: The intersection of two sets contains only the elements that are common to both.

Let the cost of one pencil be \( x \) and the cost of one pen be \( y \).


From the given conditions, we can form the following linear equations:
\[ 6x + 4y = 50 \]
\[ 5x + 6y = 46 \]

Thus, these two equations represent the given conditions in the form of linear equations.
Quick Tip: Linear equations in two variables can be represented in the form \( ax + by = c \).

Expanding the given equation: \[ (x - 2)^2 + 1 = 2x - 3 \]
\[ x^2 - 4x + 4 + 1 = 2x - 3 \]
\[ x^2 - 4x + 5 = 2x - 3 \]

Rearrange the terms: \[ x^2 - 4x + 5 - 2x + 3 = 0 \]
\[ x^2 - 6x + 8 = 0 \]

Since the equation is in the form \( ax^2 + bx + c = 0 \) where \( a = 1, b = -6, c = 8 \), it is a quadratic equation.
Quick Tip: A quadratic equation always has a variable raised to the power of 2 as its highest degree.

The general formula for finding the \( n^{th} \) term of an arithmetic progression (A.P.) is given by: \[ a_n = a + (n-1)d \]
where:

\( a_n \) represents the \( n^{th} \) term of the A.P.
\( a \) is the first term of the sequence.
\( d \) is the common difference, which is the difference between any two consecutive terms.
\( n \) is the number of terms in the sequence.

This formula helps in finding any term of an arithmetic sequence without listing all preceding terms.
Quick Tip: In an arithmetic progression, each term increases or decreases by a fixed amount called the \textbf{common difference} (\( d \)).

To calculate the distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \), we use the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substituting \( (x_1, y_1) = (7,8) \) and \( (x_2, y_2) = (-2,3) \):
\[ d = \sqrt{(-2 - 7)^2 + (3 - 8)^2} \]
\[ = \sqrt{(-9)^2 + (-5)^2} = \sqrt{81 + 25} = \sqrt{106} \]

Thus, the distance between the two points is \( \sqrt{106} \).
Quick Tip: The \textbf{distance formula} is derived from the \textbf{Pythagorean theorem} in coordinate geometry.

The given problem forms a right-angled triangle where:
- \( OQ \) is the hypotenuse (distance from the centre to point \( Q \)) = 25 cm.
- \( PQ \) is the tangent length = 24 cm.
- \( OP \) is the radius of the circle.

Using Pythagoras’ theorem: \[ OQ^2 = OP^2 + PQ^2 \]

Substituting values: \[ 25^2 = r^2 + 24^2 \]
\[ 625 = r^2 + 576 \]
\[ r^2 = 49 \quad \Rightarrow \quad r = 7 cm \]

Thus, the radius of the circle is 7 cm.
Quick Tip: The \textbf{tangent to a circle is always perpendicular to the radius} at the point of contact.

Using the Pythagorean identity:
\[ \sin^2 A + \cos^2 A = 1 \]

Substituting \( \cos A = \frac{12}{13} \):
\[ \sin^2 A = 1 - \left(\frac{12}{13}\right)^2 \]
\[ \sin^2 A = 1 - \frac{144}{169} = \frac{25}{169} \]
\[ \sin A = \frac{5}{13} \]

Now, calculating \( \tan A \):
\[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12} \]

Thus, the values are: \( \sin A = \frac{5}{13} \) and \( \tan A = \frac{5}{12} \).
Quick Tip: The \textbf{Pythagorean identity} helps find missing trigonometric values using known ratios.

A die has six faces: \( \{1,2,3,4,5,6\} \).


i) Probability of getting a prime number:

Prime numbers in this range: \( \{2,3,5\} \)

Total favorable outcomes = 3, total possible outcomes = 6.
\[ P(Prime) = \frac{3}{6} = \frac{1}{2} \]

ii) Probability of getting an odd number:

Odd numbers in this range: \( \{1,3,5\} \)

Total favorable outcomes = 3, total possible outcomes = 6.
\[ P(Odd) = \frac{3}{6} = \frac{1}{2} \]

Thus, the probability of getting a prime number is \( \frac{1}{2} \) and the probability of getting an odd number is \( \frac{1}{2} \).
Quick Tip: The \textbf{probability of an event} is calculated as: \[ P(E) = \frac{Number of favorable outcomes}{Total outcomes} \]

We start with the given equation: \[ 2 \log 5 + \frac{1}{2} \log 9 - \log 3 = \log x \]

Using the logarithmic identity \( a \log b = \log b^a \), we rewrite:
\[ \log 5^2 + \log 9^{\frac{1}{2}} - \log 3 = \log x \]
\[ \log 25 + \log 3 - \log 3 = \log x \]

Since \( \log 3 - \log 3 = 0 \), we are left with:
\[ \log 25 = \log x \]

By comparing, we get:
\[ x = 25 \]

Thus, the value of \( x \) is 25.
Quick Tip: Use logarithmic properties: \( a \log b = \log b^a \) and \( \log A + \log B = \log (A \times B) \) to simplify expressions.

To check whether \( -3 \) and \( 3 \) are zeroes of the given polynomial:
\[ f(x) = x^4 - 81 \]

Step 1: Substitute \( x = -3 \) \[ f(-3) = (-3)^4 - 81 \] \[ = 81 - 81 = 0 \]

Step 2: Substitute \( x = 3 \) \[ f(3) = (3)^4 - 81 \] \[ = 81 - 81 = 0 \]

Since \( f(-3) = 0 \) and \( f(3) = 0 \), both \( -3 \) and \( 3 \) satisfy the polynomial equation.

Thus, \(-3\) and \(3\) are zeroes of the polynomial \( x^4 - 81 \).
Quick Tip: To check if a number is a zero of a polynomial, substitute it into the equation. If the result is zero, it is a valid zero.

We use the elimination method to solve the given equations.

Step 1: Multiply both equations to make the coefficients of one variable equal.

Multiply the first equation by \(3\) and the second by \(2\) to equalize the coefficients of \( y \):
\[ (3x + 2y) \times 3 = (-1) \times 3 \quad \Rightarrow \quad 9x + 6y = -3 \]
\[ (2x + 3y) \times 2 = (-9) \times 2 \quad \Rightarrow \quad 4x + 6y = -18 \]

Step 2: Subtract the equations.
\[ (9x + 6y) - (4x + 6y) = (-3) - (-18) \]
\[ 5x = 15 \]
\[ x = 3 \]

Step 3: Substitute \( x = 3 \) into the first equation.
\[ 3(3) + 2y = -1 \]
\[ 9 + 2y = -1 \]
\[ 2y = -10 \]
\[ y = -5 \]

Thus, the solution is \( \mathbf{x = 3, y = -5} \).
Quick Tip: The elimination method is useful when the coefficients of a variable can be made equal by multiplication, allowing for easier cancellation.

Let Rohan’s present age be \( x \) years. Then his mother’s age is:
\[ (x + 26) years \]

After 3 years, their ages will be:
\[ (x + 3) and (x + 26 + 3) = (x + 29) \]

According to the given condition:
\[ (x + 3)(x + 29) = 360 \]

Expanding:
\[ x^2 + 29x + 3x + 87 = 360 \]
\[ x^2 + 32x + 87 - 360 = 0 \]
\[ x^2 + 32x - 273 = 0 \]

Thus, the required quadratic equation is:
\[ \mathbf{x^2 + 32x - 273 = 0} \] Quick Tip: To form a quadratic equation, express unknown values in terms of a single variable and use given conditions to set up an equation.

From a point outside a given circle, exactly two tangents can be drawn to the circle.

A tangent is a line that touches the circle at exactly one point without crossing it. Since an external point can have two such contact points on the circle, the total number of tangents that can be drawn is:
\[ \mathbf{2} \] Quick Tip: From any point outside a circle, exactly two tangents can be drawn to the circle.

The total surface area of a cylinder is given by the formula:
\[ Total Surface Area = 2\pi r (r + h) \]

Given: \[ Radius = \frac{Diameter}{2} = \frac{2}{2} = 1 m, \quad Height = 7 m \]
\[ Surface Area = 2\pi (1) (1 + 7) = 2\pi (8) = 16\pi m^2 \]

Approximating \( \pi = 3.14 \):
\[ Surface Area = 16 \times 3.14 = 50.24 m^2 \]

The cost per square meter is ₹3, so the cost to paint one drum is:
\[ 50.24 \times 3 = 150.72 \]

For 10 drums:
\[ 150.72 \times 10 = 1507.2 \]

Thus, the total charge to be paid is ₹ 1507.20.
Quick Tip: To find the total surface area of a cylinder, use the formula: \( 2\pi r (r + h) \).

Rewriting \( \cot^2 A \) in terms of \( \tan^2 A \):
\[ \cot^2 A = \frac{1}{\tan^2 A} \]

Substituting in the given equation:
\[ \frac{1 - \tan^2 A}{\frac{1}{\tan^2 A} - 1} \]
\[ = \frac{1 - \tan^2 A}{\frac{1 - \tan^2 A}{\tan^2 A}} \]
\[ = \tan^2 A \]

Thus, the given equation holds true.
Quick Tip: Use trigonometric identities like \( \cot A = \frac{1}{\tan A} \) to simplify expressions.

To find the mode, we use the mode formula for grouped data:
\[ Mode = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \]

where:

\( L \) = Lower boundary of the modal class
\( f_1 \) = Frequency of the modal class
\( f_0 \) = Frequency of the class preceding the modal class
\( f_2 \) = Frequency of the class succeeding the modal class
\( h \) = Class width


Step 1: Identify the modal class

The modal class is the class with the highest frequency. From the table, the highest frequency is 8, corresponding to the class 3 – 5.
\[ L = 3, \quad f_1 = 8, \quad f_0 = 7, \quad f_2 = 2, \quad h = 2 \]

Step 2: Apply the formula
\[ Mode = 3 + \left( \frac{8 - 7}{(2 \times 8) - 7 - 2} \right) \times 2 \]
\[ = 3 + \left( \frac{1}{16 - 9} \right) \times 2 \]
\[ = 3 + \left( \frac{1}{7} \right) \times 2 \]
\[ = 3 + 0.29 = 3.29 \]

Thus, the mode of the data is approximately 3.29.
Quick Tip: The modal class is the class with the highest frequency. Use the formula for grouped data to estimate the mode accurately.

We will prove this by contradiction. Assume that \( 6 + \sqrt{2} \) is a rational number.

Step 1: Express it as a rational number.

Let \( 6 + \sqrt{2} = r \), where \( r \) is rational.

Rearrange:
\[ \sqrt{2} = r - 6 \]

Since \( r \) is rational, \( r - 6 \) is also rational. However, we know that \( \sqrt{2} \) is irrational, and a rational number cannot be equal to an irrational number.

Step 2: Contradiction

Since our assumption leads to a contradiction, \( 6 + \sqrt{2} \) must be irrational.
Quick Tip: The sum or difference of a rational and an irrational number is always irrational.

To prove that \( a_n \) forms an arithmetic progression (AP), we check if the difference between consecutive terms is constant.

For \( a_n = 3 + 4n \):

First term: \( a_1 = 3 + 4(1) = 7 \)

Second term: \( a_2 = 3 + 4(2) = 11 \)

Third term: \( a_3 = 3 + 4(3) = 15 \)


Common difference:
\[ d = a_2 - a_1 = 11 - 7 = 4 \]

Since the difference is constant, it forms an AP.

Sum of first 15 terms:
\[ S_n = \frac{n}{2} (2a + (n-1)d) \]
\[ S_{15} = \frac{15}{2} (2(7) + 14(4)) \]
\[ = \frac{15}{2} (14 + 56) = \frac{15}{2} (70) = 525 \]

For \( a_n = 9 - 5n \):

First term: \( a_1 = 9 - 5(1) = 4 \)

Second term: \( a_2 = 9 - 5(2) = -1 \)

Third term: \( a_3 = 9 - 5(3) = -6 \)


Common difference:
\[ d = a_2 - a_1 = -1 - 4 = -5 \]

Since the difference is constant, it forms an AP.

Sum of first 15 terms:
\[ S_{15} = \frac{15}{2} (2(4) + 14(-5)) \]
\[ = \frac{15}{2} (8 - 70) = \frac{15}{2} (-62) = -465 \]

Thus, both sequences form an AP, and the sums of the first 15 terms are 525 and -465 respectively.
Quick Tip: In an arithmetic progression (AP), the sum of \( n \) terms is given by: \[ S_n = \frac{n}{2} (2a + (n-1)d) \]

The largest right circular cone that can be cut from a cube has its base inscribed within one face of the cube, and its height equals the edge length of the cube.

Step 1: Identify given values.
\[ Edge of the cube = 7 cm \]
\[ Radius of the base = \frac{Edge}{2} = \frac{7}{2} = 3.5 cm \]
\[ Height of the cone = 7 cm \]

Step 2: Use the volume formula for a cone.
\[ V = \frac{1}{3} \pi r^2 h \]
\[ V = \frac{1}{3} \pi (3.5)^2 (7) \]
\[ V = \frac{1}{3} \pi (12.25) (7) \]
\[ V = \frac{85.75\pi}{3} \approx 89.92 cm^3 \]

Thus, the volume of the cone is approximately 89.92 cm³.
Quick Tip: The largest cone inside a cube has a height equal to the cube’s edge and a base radius equal to half the edge.

The union of two sets \( A \) and \( B \) consists of all elements present in either \( A \) or \( B \) or both.
\[ A \cup B = \{1,2,3,4,5\} \cup \{3,4,5,6,7\} \]
\[ = \{1,2,3,4,5,6,7\} \] Quick Tip: Union (\(\cup\)) of two sets includes all unique elements from both sets.

\[ B \cup C = \{3,4,5,6,7\} \cup \{1,3,5,7\} \]
\[ = \{1,3,4,5,6,7\} \] Quick Tip: The union of two sets combines all elements, removing any duplicates.

\[ A \cup D = \{1,2,3,4,5\} \cup \{2,4,6,8\} \]
\[ = \{1,2,3,4,5,6,8\} \] Quick Tip: The union operation does not repeat elements. It only includes distinct values.

The difference \( B - D \) consists of elements in \( B \) that are not in \( D \).
\[ B - D = \{3,4,5,6,7\} - \{2,4,6,8\} \]
\[ = \{3,5,7\} \] Quick Tip: The difference (\(-\)) between two sets includes elements present in the first set but not in the second.

The intersection \( A \cap B \) consists of elements that are common in both \( A \) and \( B \).
\[ A \cap B = \{1,2,3,4,5\} \cap \{3,4,5,6,7\} \]
\[ = \{3,4,5\} \] Quick Tip: The intersection (\(\cap\)) operation finds elements that are present in both sets.

\[ B \cap D = \{3,4,5,6,7\} \cap \{2,4,6,8\} \]
\[ = \{4,6\} \] Quick Tip: The intersection of two sets contains only the elements that are found in both sets.

\[ C \cap D = \{1,3,5,7\} \cap \{2,4,6,8\} \]
\[ = \emptyset \]

Since there are no common elements, the result is the empty set \( \emptyset \).
Quick Tip: If two sets have no common elements, their intersection is the empty set (\(\emptyset\)).

The difference \( A - D \) consists of elements in \( A \) that are not in \( D \).
\[ A - D = \{1,2,3,4,5\} - \{2,4,6,8\} \]
\[ = \{1,3,5\} \] Quick Tip: Set difference (\(-\)) includes elements in the first set that are not in the second set.

To find the median class, we use the following steps:

Step 1: Compute cumulative frequency (CF)

Step 2: Identify the median class

The total number of students = \( 30 \). The median class corresponds to the cumulative frequency just greater than \( \frac{30}{2} = 15 \).
\[ Median class = 55 - 60 \]

Step 3: Apply the median formula
\[ Median = L + \left( \frac{\frac{n}{2} - CF}{f} \right) \times h \]

where:


\( L = 55 \) (Lower boundary of median class)
\( n = 30 \) (Total frequency)
\( CF = 13 \) (Cumulative frequency before median class)
\( f = 6 \) (Frequency of median class)
\( h = 5 \) (Class width)

\[ Median = 55 + \left( \frac{15 - 13}{6} \right) \times 5 \]
\[ = 55 + \left( \frac{2}{6} \times 5 \right) \]
\[ = 55 + \left( \frac{10}{6} \right) \]
\[ = 55 + 1.67 = 56.67 \]

Thus, the median weight is approximately 56.67 kg.
Quick Tip: To find the median class, locate the class interval where the cumulative frequency exceeds \( \frac{n}{2} \).

Three points are collinear if the area of the triangle formed by them is zero. The area of a triangle formed by points \( (x_1, y_1) \), \( (x_2, y_2) \), \( (x_3, y_3) \) is given by:
\[ Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting the given points:
\[ \frac{1}{2} \left| 1(b - (-4)) + 3(-4 - 2) + 5(2 - b) \right| = 0 \]
\[ \frac{1}{2} \left| 1(b + 4) + 3(-6) + 5(2 - b) \right| = 0 \]
\[ \frac{1}{2} \left| (b + 4) - 18 + 10 - 5b \right| = 0 \]
\[ \frac{1}{2} \left| -4b - 4 \right| = 0 \]
\[ \left| -4b - 4 \right| = 0 \]
\[ -4b - 4 = 0 \]
\[ -4b = 4 \]
\[ b = -1 \]

Thus, the value of \( b \) for which the points are collinear is -1.
Quick Tip: Three points are collinear if the area of the triangle formed by them is zero.

Let the initial distance of the boy from the temple be \( x \) meters and the distance he walked be \( d \) meters.

The height of the temple above the boy’s eye level:
\[ Effective height = 30 - 1.5 = 28.5 m \]

Using the tan function in the right-angled triangles:
\[ \tan 30^\circ = \frac{28.5}{x + d}, \quad \tan 60^\circ = \frac{28.5}{d} \]

Substituting values:
\[ \frac{28.5}{x + d} = \frac{1}{\sqrt{3}}, \quad \frac{28.5}{d} = \sqrt{3} \]

Solving for \( x + d \) and \( d \):
\[ x + d = 28.5\sqrt{3}, \quad d = \frac{28.5}{\sqrt{3}} \]
\[ d = \frac{28.5 \times \sqrt{3}}{3} = 9.5\sqrt{3} \]

Approximating:
\[ d \approx 9.5 \times 1.732 = 16.45 m \]

Thus, the boy walked approximately 16.45 meters towards the temple.
Quick Tip: Use trigonometric ratios such as tan for height-distance problems involving right triangles.

The total number of cards in a deck = 52.

i) Probability of getting a king of red color:
There are 2 red kings (King of hearts and King of diamonds).
\[ P(Red King) = \frac{2}{52} = \frac{1}{26} \]

ii) Probability of getting a face card:
There are 12 face cards (4 kings, 4 queens, 4 jacks).
\[ P(Face Card) = \frac{12}{52} = \frac{3}{13} \]

iii) Probability of getting a jack of hearts:
There is only one jack of hearts in a deck.
\[ P(Jack of Hearts) = \frac{1}{52} \]

iv) Probability of getting a spade:
There are 13 spades in a deck.
\[ P(Spade) = \frac{13}{52} = \frac{1}{4} \] Quick Tip: Probability is calculated as: \[ P(E) = \frac{Favorable outcomes}{Total outcomes} \]

Step 1: Construct the given triangle.
1. Draw a base of 6 cm.
2. From each endpoint of the base, draw two arcs of 5 cm radius and mark the intersection as the third vertex.
3. Join all points to form the isosceles triangle.

Step 2: Construct the similar triangle.
1. Draw a ray from one vertex making an acute angle with the base.
2. Mark three equal segments (since \(3 > 2\) in \( \frac{2}{3} \)).
3. Connect the second division point to the original vertex.
4. Draw a parallel line to the opposite side of the triangle using a compass.
5. The new triangle is similar to the original one.


Quick Tip: To construct a similar triangle, divide the base into equal parts according to the given ratio.

To find the zeroes of the polynomial \( p(x) = x^2 - 3x - 4 \), we solve:
\[ x^2 - 3x - 4 = 0 \]

Step 1: Factorize the equation.
\[ (x - 4)(x + 1) = 0 \]

Solving:
\[ x - 4 = 0 \Rightarrow x = 4, \quad x + 1 = 0 \Rightarrow x = -1 \]

Thus, the zeroes of the polynomial are \( x = 4 \) and \( x = -1 \).

Step 2: Plot the graph.
1. Construct a table of values for \( y = x^2 - 3x - 4 \).
2. Plot points and draw a parabolic curve.
3. The points where the curve crosses the x-axis give the zeroes at \( x = 4 \) and \( x = -1 \). Quick Tip: The zeroes of a polynomial correspond to the x-values where the curve intersects the x-axis.